reality of motion - Page 2

Guille · 04-17-2005, 02:51 PM

Quote:

Originally Posted by AntonioLao

the square root of a number (x) has two solutions given by the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

if x is expressed by the algebraic polynomial equation of 2nd degree.

$ax^2+bx+c=0$

it was the argument based on these two solutions that Dirac postulated the existence of antimatter but his math used the concept of spinor algebra.

How did he get to the same result from a different way? do you know what exactly did he do?

AntonioLao · 04-18-2005, 12:11 PM

GUILLE,

I have to look up Dirac's method before replying this post.

antonio

AntonioLao · 04-18-2005, 02:26 PM

Quote:

Originally Posted by GUILLE

How did he get to the same result from a different way?

Reference: Schiff, "Quantum Mechanics," 2nd Ed., Chap XII, p323, p338, McGraww-Hill, 1955.

Both the Schroedinger and Dirac relativistic equations have negative solutions for kinetic energy and rest mass. These solutions is the same as the negative square root of the classical energy equation from special relativity $E=\pm\sqrt{c^2\mathbf{p}^2+m^2c^4}$ .

Guille · 04-22-2005, 01:05 PM

Quote:

Originally Posted by AntonioLao

in terms of m and c, quantized space H is given by

$H=m^2C^4$

this can only be true if the total linear momentum of the universe is zero and m is the rest mass of the universe, which include antimatter, dark matter, and matter.

So, E energy equals the square root of continous space.

Then, if E^2=m^2c^4 from which we get E=mc^2 is this correct:

square root of E=square root of m times c??

AntonioLao · 04-22-2005, 01:25 PM

Quote:

Originally Posted by GUILLE

square root of E=square root of m times c??

According to the strict rules of mathematics, not all functions necessarily have inverse transformations and for a matrix, if the determinant is zero, the matrix is singular and it has no inverse. The existence of an inverse is similar to the existence of a divisor such that matrix product of a matrix and its inverse is equal to the identity matrix.

$M \times M^{-1} = I$

~~subversion~~ · 04-22-2005, 02:31 PM

Quote:

Originally Posted by AntonioLao

in terms of m and c, quantized space H is given by

$H=m^2C^4$

this can only be true if the total linear momentum of the universe is zero and m is the rest mass of the universe, which include antimatter, dark matter, and matter.

I am curious, what do you mean by linear momentum of the universe?

Guille · 04-22-2005, 02:34 PM

Quote:

Originally Posted by subversion

I am curious, what do you mean by linear momentum of the universe?

he doesn't mean the linear momentum of the universe. bu the total linear momentum in the unvierse. your momentum, mine, the earth's.....don't you, AL?

AntonioLao · 04-22-2005, 02:48 PM

I'm trying to prove that the total linear momentum of the universe is zero. If the total linear momentum is zero, the universe as a whole has no motion. It's not going anywhere. It is here and there forevermore. I see you but I can't get there because i'm always here (an everlasting consciousness even beyond the physical death).

Guille · **It might be moving, but...... -** 04-23-2005, 03:54 AM

If you don't manage to, or nobody does, then it wouldn't be such a problem, the universe is moving, and? we can still live.

The problems that appear in my head with the universe moving, or having linear momentum, are the same as imagening the universe growing, and it is, were is it moving/growing? what is it mayde of there? is the thing out there inside a thing even more and more out there?................and questions of that type.

maybe our universe has circular momentum and it is going around a really really super-mega-hyper-ultra star which atracts it and many other universes with a mega-gravitational field. no?

Guille · 04-23-2005, 04:02 AM

Quote:

Originally Posted by AntonioLao

These solutions is the same as the negative square root of the classical energy equation from special relativity $E=\pm\sqrt{c^2\mathbf{p}^2+m^2c^4}$ .

what is the P their?